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nirvana33
16 days ago
6

A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th

e computer. Consider a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions
(5 min, sec 1.10)By how much is the total time reduced if the time for FP operations is reduced by 20%?


(5 min, sec 1.10) By how much is the time for INT operations reduced if the total time is reduced by 20%?
(5 min, sec 1.10) Can the total time be reduced by 20% by reducing only the time for branch instructions?
Engineering
1 answer:
alex41 [274]16 days ago
8 0

Answer:

a) In this scenario, the time required for the FP operation would decrease by 20%, which means it would now take 80% of the original duration.

(1-0.2)*70 s =56s

The reduction in this case is 70-56 s=14s

And the new total time would be calculated as 250-14=236 s

b) Here, the overall time is reduced by 20%, indicating that the updated total time would amount to 0.8 times the original total time (1-0.2) *250s =200 s

The initial duration for INT operations can be determined as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

For this segment, it is assumed that the INT operations time is the only one adjusted:

200 = 70+85+40 \Delta t_{INT}

And subsequently: \Delta t_{INT}= 200-70-85-40=5 s

c) A decrease in overall time suggests the new total would stand at 205 s based on preceding calculations. Timing for FP is now 70 s, for L/S is 85 s, and for INT operations is 55 s. Thus, summing them gives 70+85+55=210 s, indicating that reducing branch instruction times alone cannot lead to a 20% reduction in overall time.

Explanation:

The information shows that a program on a computer is run for 250 s, with 70 s allocated to FP instructions, 85 s for L/S instructions, and 40 s for branch instructions.

Part 1

In this scenario, the required FP operation time diminishes by 20%, indicating it would occupy 80% of the original time.

(1-0.2)*70 s =56s

The reduction in this instance equates to 70-56 s=14s

And the new total time is given by 250-14=236 s

Part 2

<pfor this="" case="" we="" denote="" the="" total="" time="" reduction="" at="" which="" correlates="" to="" a="" new="" of="" times="" original="" cumulative="" src="https://tex.z-dn.net/?f=%20%281-0.2%29%20%2A250s%20%3D200%20s" id="TexFormula12" title=" (1-0.2) *250s =200 s" alt=" (1-0.2) *250s =200 s" align="absmiddle" class="latex-formula">

The original INT operations timing is detailed as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

In this context, we assume the modification pertains solely to INT operations:

200 = 70+85+40 \Delta t_{INT}

And following that: \Delta t_{INT}= 200-70-85-40=5 s

Moreover, we can quantify the decline using the proportional change:

\% Change = \frac{5s}{55 s} *100 = 9.09\% in reduction.

Part 3

<pthe decrease="" in="" total="" time="" signifies="" it="" would="" be="" s="" derived="" from="" the="" above="" calculations.="" given="" times="" of="" for="" fp="" l="" and="" int="" operations="" results="" illustrating="" that="" merely="" reducing="" branch="" instruction="" will="" not="" enable="" a="" cut="" time.=""> </pthe></pfor>
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